Question

A differential equation and a nontrivial solution f are given below. Find a second linearly independent solution using reduction of order. Assume that all constants of integration are zero.
tx" - (3t + 1)x' + 3x = 0; t>0, f(t) = 4e³ᵗ

Answer 1

Answer:

The second linearly independent solution is

g(t) = = -(4/9)(3t + 1)

Step-by-step explanation:

Given the differential equation

tx'' - (3t + 1)x' + 3x = 0 ...................(1)

and a solution

f(t) = 4e^(3t)

We want to find a second linearly independent solution g(t), using the method of reduction of order.

Let this second solution be

x = uf(t)

x = u. 4e^(3t) ...................................(2)

x' = u'. 4e^(3t) + u. 12e^(3t) ...........(3)

x'' = u''. 4e^(3t) + u'. 12e^(3t) + u'. 12e^(3t) + u. 36e^(3t)

= 4u''e^(3t) + 24u'e^(3t) + 36ue^(3t) .............(4)

Using the values of x, x', and x'' in (2), (3), and (4) in (1), we have

t[4u''e^(3t) + 24u'e^(3t) + 36ue^(3t)] - (3x + 1)[u'. 4e^(3t) + u. 12e^(3t)] + 3u. 4e^(3t) = 0

4tu''e^(3t) + (12t - 4)u'e^(3t) = 0......(5)

Let w = u'

then w' = u''

(5) now becomes

4tw'e^(3t) + (12t - 4)we^(3t) = 0

4tw'e^(3t) = (4 - 12t)we^(3t)

w'/w = (4 - 12t)/4t = (1/t) - 3

Integrating this, assuming all constants of integration are 0, we have

lnw = lnt - 3t

w = e^(lnt - 3t) = e^(lnt)e^(-3t)

w = te^(-3t)

But remember w = u'

=> u' = te^(-3t)

Integrating this, by part, taking constant of integration as 0, we have

u = -(1/9)(3t + 1)e^(-3t)

Using this in (2)

x = 4 [-(1/9)(3t + 1)e^(-3t)]e^(3t)

= -(4/9)(3t + 1)

And this is what we are looking for.